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Irina18 [472]
3 years ago
7

Symptoms of excessive stress include all of the following EXCEPT:

Physics
1 answer:
sweet-ann [11.9K]3 years ago
6 0

Answer:

Symptoms of excessive stress include all of the following EXCEPT: increased energy level.

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Which are properties of a liquid? Check all that apply.
Strike441 [17]

Answer:

the property of liquid are

1 they can flow from one place to another if surface is slanted

2 it cannot be compressed

6 0
3 years ago
Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat
kvasek [131]

Answer:

a) a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

b) k = \frac{9.8}{9.74}=1.006

Explanation:

Part a

For this case we can begin finding the period like this:

T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s

Then we know that the centripetal acceleration is given by:

a= \frac{v^2}{r}

And the velocity is given by:

v=\frac{2\pi r}{T}

If we replace this into the acceleration we got:

a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}

And we can replace the values and we got:

a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

Part b

For this case we want to find a value of k such that:

a= k 9.8

Where a = 9.74, so then we can solve for k like this:

k = \frac{9.8}{9.74}=1.006

8 0
3 years ago
A solenoid field is:
Bumek [7]

increased with an increased current flow

7 0
3 years ago
The mass of a moving object increases, but its speed stays the same. What happens to the kinetic energy of the object as a resul
ozzi
The correct answer is 
B It increases.

In fact, the kinetic energy of a moving object is given by
K= \frac{1}{2}mv^2
where m is the mass of the object and v is its speed. We see that the kinetic energy is proportional to the mass and proportional to the square of the speed: in this problem, the speed of the object remains the same, while its mass increases, therefore the kinetic energy will increase as well.
5 0
4 years ago
Read 2 more answers
Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o
Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

\Sigma F = F - F' = M\cdot a

Box with mass 2M

\Sigma F = F' - F'' = 2\cdot M \cdot a

Box with mass 3M

\Sigma F = F'' = 3\cdot M \cdot a

On the third equation, acceleration can be modelled in terms of F'':

a = \frac{F''}{3\cdot M}

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

F' = 2\cdot M \cdot a + F''

F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''

F' = \frac{5}{3}\cdot F''

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)

F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''

F = 2\cdot F''

F'' = \frac{1}{2}\cdot F

Afterwards, F' as function of the external force can be obtained by direct substitution:

F' = \frac{5}{6}\cdot F

The net forces of each block are now calculated:

Box with mass M

M\cdot a = F - \frac{5}{6}\cdot F

M\cdot a = \frac{1}{6}\cdot F

Box with mass 2M

2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F

2\cdot M \cdot a = \frac{1}{3}\cdot F

Box with mass 3M

3\cdot M \cdot a = \frac{1}{2}\cdot F

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

8 0
4 years ago
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